Research question: Does happiness differ between the Swiss regions?
2022/09/29
Research question: Does happiness differ between the Swiss regions?
Variables:
Nuts2: Large RegionsH1 (variable name not a hypothesis): Q1 How happy or unhappy [1 Completely happy - 7 Completely unhappy]H1 is obviously ordinal - can mean even be appropriate?
Hypothesis 1: The respondents from the 7 regions reported different mean happiness levels.
Hypothesis 2: Respondents from Espace Mittelland reported higher mean happiness levels than Zentralschweiz.
| Nuts2 | n | mean | trimmed10 | median | sd | var | skew | kurt |
|---|---|---|---|---|---|---|---|---|
| Région lémanique | 570 | 2.91 | 2.91 | 3 | 0.92 | 0.85 | 0.17 | 3.58 |
| Espace Mittelland | 720 | 2.72 | 2.72 | 3 | 0.89 | 0.79 | 0.36 | 3.01 |
| Nordwestschweiz | 427 | 2.70 | 2.70 | 3 | 0.88 | 0.78 | 0.47 | 4.05 |
| Zürich | 513 | 2.76 | 2.76 | 3 | 0.97 | 0.95 | 0.59 | 3.93 |
| Ostschweiz | 433 | 2.73 | 2.73 | 3 | 0.95 | 0.90 | 0.75 | 4.56 |
| Zentralschweiz | 275 | 2.61 | 2.61 | 3 | 0.85 | 0.73 | 0.62 | 4.00 |
| Ticino | 160 | 2.91 | 2.91 | 3 | 0.98 | 0.95 | 1.16 | 6.14 |
Box plots are excellent to display distributions.
Why are they not a good choice in case?
WARNING: depending on the bin size histograms can be misleading.
Quantile-Quantile-plots are a great way to compare the sample distribution to a theoretical distribution. Ideally, the points would match the line.
Why do we see a stair pattern?
oneway.test(H1~Nuts2,var.equal=FALSE, data=df_1f)
## ## One-way analysis of means (not assuming equal variances) ## ## data: H1 and Nuts2 ## F = 5.2507, num df = 6.0, denom df = 1030.9, p-value = 2.434e-05
Levine and Hullett (2002) recommend Ω² or η² as effect size for ANOVAs.
aov(H1~Nuts2, data=df_1f) %>% effectsize::omega_squared(verbose=F) %>% toTable()
| Parameter | Omega2 | CI | CI_low | CI_high |
|---|---|---|---|---|
| Nuts2 | 0.0079776 | 0.95 | 0.002278 | 1 |
Hypothesis 1: The respondents from the 7 regions reported different mean happiness levels. –> Null-Hypothesis can be rejected, but the effect is minimal
f1_lm <- lm(H1~Nuts2, data=df_1f)
f1_emm <- emmeans::emmeans(f1_lm, 'Nuts2', data=df_1f)
emmeans::test(
emmeans::contrast(
f1_emm,
list(ac1=c(0, 1, 0, 0, 0, -1, 0)) # this list can contain multiple contrasts
),
adjust='none'
)
## contrast estimate SE df t.ratio p.value ## ac1 0.114 0.0651 3091 1.753 0.0797